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Calculate $\lim_{n\to\infty}\hspace{2 pt}a_n$, where $a_n = \sqrt{n^2+n} - n$ is a sequence of complex numbers.

I performed the ratio test:

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{\sqrt{(n+1)^2 + (n+1)} - (n+1)}{\sqrt{n^2+n} - n} = \lim_{n\to\infty}\frac{\sqrt{n^2+3n+2} - (n+1)}{\sqrt{n^2+n} - n}$$

which ultimately leads to, when multiplied by $\frac{\frac{1}{n}}{\frac{1}{n}}$,

$$\lim_{n\to\infty}\frac{\sqrt{1+\frac{3}{n}+\frac{2}{n^2}} - (1+\frac{1}{n})}{\sqrt{1+\frac{1}{n}} - 1}$$

Should I multiply by the complex conjugate? It leads to a rather messy expression. I am stuck.

How can one perform the root test here? I attempted: $$\lim_{n\to\infty}\sqrt[n]{\sqrt{n^2+n} - n} \Longleftrightarrow \lim_{n\to\infty}((n^2 + n)^{\frac{1}{2}} - n)^{\frac{1}{n}}$$

I have also no idea how to proceed.

Martin Sleziak
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Don Larynx
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2 Answers2

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$$a_n =\sqrt{n^2+n}-n = \frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} \Rightarrow \lim_{n \to \infty} a_n = \frac{1}{2}$$

aaaaa
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2

$$a_n = \sqrt{n^2+n} - n = \dfrac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} \cdot (\sqrt{n^2+n}-n) = \dfrac{n^2+n - n^2}{\sqrt{n^2+n}+n} = \dfrac{n}{\sqrt{n^2+n} + n} = \dfrac1{\sqrt{1+1/n}+1}$$ Now let $n \to \infty$ to conclude the limit.